WEBVTT mathematics/probability/murray
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Hello, you are watching the probability lectures here on www.educator.com, my name is Will Murray.
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Currently, we are working our way through the continuous distributions.
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We have already had videos on the uniform distribution and the Gamma distribution
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which included the exponential distribution, and the Chi square distribution.
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Today, we are going to talk about the last of the continuous distributions which is the β distribution.
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It is similar to the Gamma distribution, you will see some of the same elements but of course, it is also different.
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Let us check it out.
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Before we jump into the actual β distribution, we have to learn what the β function is.
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The first thing that you have to keep straight here is that, there is a β function and then there is β distribution.
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They are not exactly the same thing.
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We are going to use the β function in the process of defining the β distribution.
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In that sense, it is like the Gamma distribution where we had a Gamma function and then we also had a Gamma distribution.
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The Gamma function was just one part of the Gamma distribution.
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Let us see what the β function is.
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We had two fixed parameters, there is always an Α and there is a β.
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The β distribution and the β function, it is the whole family of things.
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Because you could pick any different value for α you want and any different value for β you want.
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Let us define the β function.
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What we do is, you want to think of Α and β as being numbers.
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You take these numbers α and β, and you plug them into this integral.
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We have the integral of Y ^α – 1 × 1 – Y ^β – 1 DY, that is just some integral.
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Remember, α and β are constants.
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We plug in constant values and then, you have an integral that
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you can solve using calculus 2 methods and you will get some constant number.
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The idea is that you plug in Α and β, the β function will spit out a particular number, a constant value.
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There is a relationship between the Γ and the β functions,
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which often makes it quite easy to evaluate the β function, which is that B of Α × β is equal to Γ of Α × Γ β ÷ Γ of Α + β.
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That is very convenient if you have whole numbers because this relationship between Γ of N and the factorial function.
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Γ of a whole number is exactly equal to N -1!.
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If you know the numbers for Α and β here, you can just drop them into these 3 Gamma functions,
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and evaluate those using factorials, and it turns out to be something fairly easy to evaluate.
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You can find B of α and β fairly quickly as a number, just by calculating out some factorials.
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If they are not whole numbers then it is a much more difficult proposition.
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We are not going to get into that.
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Remember, this is just the β function, I have not talked yet about the β distribution.
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β function is just something we plug in two numbers, Α and β, and it spits out a number as an answer.
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Next to a step is to see how that is incorporated into the β density function,
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which is the density function for the β distribution.
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Now, we are going to talk about the β distribution.
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Remember, we already talked about the β function, that is different from the β distribution but
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it is part of the β distribution.
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The idea here is that, we want to define the double polynomial distribution.
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It is always on the interval from 0 to 1, we always have Y going from 0 to 1.
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We want to define this double polynomial distribution using essentially the function Y ^α -1 and 1 - Y ^β -1.
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The point there is that, it is symmetric between 0 and 1.
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Whatever the function does at 0, the 1 - Y behaves similarly at 1, depending on what the values of α and β are.
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That is the basic function that we want to look at.
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The problem is that, if we integrate that, we might not necessarily get it exactly equal to 1.
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The probability density function must always satisfy the property that, when you integrate it,
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we take the integral over the whole domain with the answer has to come out to be 1.
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In order to fix that, what we do is we take this Y ^α -1 and 1 – Y ^β-1.
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We just divide by the value of the integral, in order to make the integral come out to be 1.
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That is how we create the density function for the β distribution.
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We start out with Y ^α -1 × 1 – Y ^β -1.
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And then, we divide it by the integral of that function, in order to not make the total integral come out to be 1.
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You really want to think of this B of Α β in the denominator here,
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it is just a correction term that we put in there to make the total integral come out to be 1.
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It is just a constant and it is just sort of a fudge factor really,
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that is probably the best way to think of it as a fudge factor to make the integral come out to be 1,
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the integral of F of Y DY equal to 1.
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It is not really the most important part of the function here.
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The most important part are these two polynomial terms, the Y ^α -1 and 1 – Y ^β -1.
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You want to think of those as the really important part of the function.
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Those are the ones that give its shape and we will explore some of the graphs later.
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This denominator is just a constant, it is a fudge factor the gets thrown in there,
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in order to make the total integral come out to be 1.
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The denominator is exactly the β function that we learn on the previous slide.
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It is read up exactly to be the integral of Y ^α -1, 1 –Y ^β -1.
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We divide by that constant, the whole integral now comes out to be 1.
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That is the density function for the β distribution.
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Remember, it is a kind of a polynomial thing and also remember that,
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the density function and the β function are two different things, let us try to keep those straight.
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We have the β distribution, we should figure out the key properties, I have listed them here.
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Remember, the mean is always the same as expected value.
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Expected value and mean are synonymous, they mean the same thing.
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The mean for the β distribution is Α/Α + β.
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That is a fairly easy formula to keep track of.
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The variance is much more complicated and much harder to remember.
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The variance turns out to be Α × β ÷ α + β² × Α + β + 1.
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Much messier formula for the variance, we will see an example of that in the problems later on.
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You will see how that gets used.
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The standard deviation is usually not worth remembering, because you can always figure it out,
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if you remember the variance.
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The standard deviation is always the square root of the variance.
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That is true for any distribution not just the β distribution.
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The standard deviation, what we do is we just take that complicated formula for the variance
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and we slap a square root around everything.
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It is really not that enlightening by itself, it is more useful to remember the mean and variance of the β distribution.
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In example 1, we are going to calculate B of 3, 4.
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Just little practice with the β function first, before we jump into actually solving any probability problems.
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The solution here is to, there are two ways you can do it.
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One way would be to solve this as an integral.
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Here is α, and here is β.
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You could solve this out as an integral, the integral from 0 to 1 of Y ^α – 1 1 – Y ^β-1 DY.
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Because that is the definition of B of Α β, that is the definition there.
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You can solve out this integral, it would not be too bad because you plug in Α = 3 and β = 4.
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You just do a little calculus and it would come out to be some number.
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I'm not going to do it that way because I do not want to do that much calculus.
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Let me show you another method to do it, instead.
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The other method is, to remember this relationship between the β function and the Gamma function.
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We learn that several slides ago, you can go back and check that if you did not pick up on that and register it the first time.
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But, the relationship between the β function and the Gamma function is that, B of α and β is Γ of Α × Γ of β ÷ Γ of Α + β.
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That is where it translates everything here into Gamma function.
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I will go ahead and fill in α = 3 and β = 4.
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Our Gamma of α + β will be Γ of 3 + 4 is 7.
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I have to solve some Gamma functions but remember that Gamma function is,
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it can be thought of as a generalization of the factorial function.
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Γ of N is just N -1!.
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This is, Γ of 3 would be 2!, Γ of 4 would be 3!, and Γ of 7 would be 6!.
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Now, it is very easy just to calculate those factorials.
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2! Is just 2, 3! Is 6, 6! is 9 × 2 × 3 × 4 × 5 × 6.
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I'm writing it like that because now, I can cancel some things, I will not have to multiply by big numbers.
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I will cancel the 6 there and I will cancel that 2 with that 2.
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And now, I have got 1/3 × 4 × 5 which is 3 × 4 is 12 × 5 is 60, 1/60.
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To remind you how that worked.
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There are two ways I could have solve this.
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I could use the original definition of the β function which is an integral formula.
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And then, I would just plug in the values of the Α and β and done some calculus to work out the integral formula.
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I see that with the benefit of having gotten the other way, that my integral better has worked out to be 1/60.
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The way I actually calculated it, was to use this relationship between
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the β function and Gamma function which I gave you a couple slides ago.
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Translates a β function into three computations using Gamma functions.
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I dropped in the values for Α and β, and converted those into factorials.
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Remember, the Gamma function is just like the factorial function except offset by 1,
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when you have a whole number in there.
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I plugged in those factorials, calculated it out, and simplified it down to a relatively nice fraction there.
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In example 2, we are asked to graph several density functions for the β distribution,
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using several combinations of Α and β.
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Let me first remind you what the density function is for the β distribution.
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The density function for the β distribution is F of Y is equal to Y ^α – 1 1- Y ^ β -1.
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Those are really the important factors in the density function for the β distribution.
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There is one other factor but it is just a constant, and it is the sort of correction term.
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This fudge factor that gets introduced, in order to make the total integral be 1.
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I'm just going to write that as a constant here.
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I’m not going to even bother to work that out for the different combinations here because
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that is not so important to determining the shape.
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What is really important to determining the shape is the values of Α and β,
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and those factors of Y ^α -1 and 1 - Y ^β -1 in the numerator.
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Let me set up some axis here and we will take a look at these different combinations of Α and β.
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Here are my axes.
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Remember, the β distribution is always defined from Y equal 0 to Y = 1.
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Up at Y here, it is a little confusing on the X axis.
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But here is Y = 0 and here is Y = 1.
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The important thing to look at here, are the different values of Α and β.
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In particular, whether they are greater than 1 or less than 1.
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Here is why, let us look at the key part of the density function here is Y ^α – 1.
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Let us think about what that does for different values of Α.
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At Y = 0, let us think about what that does.
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If Α is bigger than 1, that means our exponent Y ^α -1 is positive.
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Y ^α-1 will go to 0.
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If Α is equal to 1 then Y ^Α -1 would just be Y at 0 go to 1.
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If Α is less than 1 then Y ^α -1 will be Y to a negative number.
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0 to negative power, it is like trying to divide by 0, that will go to positive infinity.
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That helps me characterize the behavior at Y = 0 depending on the different values of Α.
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The same kind of thing happens, if we look at 1 - Y ^β -1.
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This similar kinds of phenomenon will occur at Y = 1, because if Y goes to 1 then 1 - Y goes to 0.
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If β is bigger than 1 then 1 - Y ^β -1 goes to 0.
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If β is equal to 1 then 1 - Y ^β -1 goes to 1.
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If β is less than 1 then 1 - Y ^β -1 goes to infinity, because we are trying to take 0 to a negative exponent there.
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Now, we will look at the different combinations of α and β.
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In the first combination here, I see I got an Α of ½ and a β of 2.
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Α of ½ is less than 1 that means it is going to go to infinity at Y = 0.
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A β of 2 is bigger than 1, that means that Y = 1, it is going to go to 0.
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I got something that goes to infinity as Y goes to 0.
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And then, it comes down, let me make it a little steeper and in it is declined
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because we are only allowed at the total area 1 here.
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We are allowed to have total area 1.
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It has to come down and hit 0 at Y = 1.
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What we got right there is the graph of Α = ½ and β is equal to 2.
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Let me do the next one in a different color.
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I will do 1 and 4 in green, that Α is 1 and β is 4.
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I see when α is equal to 1 and Y is equal to 0, it goes to 1.
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Now, that is a little bit misleading.
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I cannot say it is exactly equal to 1 because I'm kind of ignoring the effect of this constant here.
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It is going to be 1 ÷ some constant.
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Let me just show it at some finite value.
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The important thing there is, it is not going 0 and it is not going to infinity.
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I see that, as it goes to Y = 1, the β value is still bigger than 1.
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It still going to go down to 0 there.
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Let me make that a little more rounded, more curved, it is not a completely straight line.
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That is my combination for Α is equal to 1 and β is equal to 4.
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And that means that, at 0 it is going to go to a finite limit and at 1 it is going to go down to 0.
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I will do the next one in red, α is 10 and β is 2.
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What is that mean, Α is 10, that is way bigger than 1 which means at Y = 0, it is definitely going to go to 0.
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It is so much bigger than 1 that, that term is going to sort of drag it down for a long time.
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Here it is at 0, at Y = 0 and it is going to drag along near 0 for quite a long time.
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Since β is 2, that means when it gets to the right hand and point here, it is still going down to 0.
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It does have to have area equal to 1, the area underneath the curve has to be equal 1 because all of these functions do.
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At some point, it is going to get some area.
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Let me show it getting some area right at, as it approaches 1 there.
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It is skewed to the right hand side, that Α = 10 drags it down on the left.
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The β = 2 does drag it down at Y = 1 but we have to have an area of 1 in there, somewhere.
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That is my 10, 2, α = 10 and β = 2.
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I’m going to try out my new purple marker for that.
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Here is α and that looks a lot like a blue to me.
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Α is 1.1 and β is 2.
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What that means, α is 1.1 and it is going to behave very similarly to the Α = 1,
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which means it is going to trying to go into a finite limit.
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Β = 2 means it is also going to be tied down at 0.
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Let me show this graph that sort of trying to go a finite limit, as it approaches Y is equal to 0.
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There it is, as it approaches Y = 0, it is trying to go to a positive limit.
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What happens is, when it gets right down to Y = 0, that 1.1 is still bigger than 1 which means it is forced to go to 0.
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Here is this graph that sort of trying to go to a finite limit and then at the last moment,
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it has to turn sharply downwards and go to 0.
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This purple graph, if you can tell the difference between the purple and the blue,
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this purple graph is Α = 1.1 and β is equal to 2.
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Let me recap all the different of things we are exploring here.
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I have to highlight them in yellow, as I go along.
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The important thing is to remember the general form of the β density function.
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Here it is right here, Y ^α-1 1 - Y ^β -1.
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And then, that constant in the denominator is really not important.
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I’m not going to worry about it, it does not affect the shape of the graph.
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There are two sub components there, the Y ^Α -1 determines what happens at Y = 0.
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You want to look at the value of Α to determine what happens at Y= 0.
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If Α is bigger than 1, then you got a positive exponent on Y, it is going to go to 0.
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If α is equal to 1, that terms drops out and it goes to a finite limit.
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If α is less than 1, then you got 0 to a negative number which means you are trying to divide by 0,
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which means you are going to go to infinity.
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If you look at the first graph here, there is an Α less than 1 and that is why it is going to infinity.
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The second graph, α is equal to 1, goes to a finite limit.
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The third graph, Α is much bigger than 1 which is why it starts off at 0 and stays at 0 for a long time.
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Finally, the 4th graph, Α is bigger than 1 but very close to 1, that is why it is sort of
00:23:33.900 --> 00:23:40.500
trying to go to a finite limit and then at the last minute, it has to drop down to 0.
00:23:40.500 --> 00:23:46.300
The second component of this function is the 1 - Y ^β -1.
00:23:46.300 --> 00:23:54.400
That determines what the graph does at Y = 1, in very much asymmetric fashion to the Y = 0,
00:23:54.400 --> 00:23:57.600
except that it is looking at the value of β, instead.
00:23:57.600 --> 00:24:01.200
If β is bigger than 1, you got something going to 0.
00:24:01.200 --> 00:24:07.200
If β is equal to 1, you got something going to 1, and or at least a constant.
00:24:07.200 --> 00:24:15.100
If β is less than 1, then you are trying to divide by 0, you are going to go to infinity again.
00:24:15.100 --> 00:24:25.900
In this case all, of the β that we looked at were all bigger than 1, which means all of these graphs sort of tied down to 0 at Y = 1.
00:24:25.900 --> 00:24:34.900
That is why all these graphs tie down here, but they all exhibit different behavior on their way to getting to that point,
00:24:34.900 --> 00:24:40.800
which is why we get this sort of interesting of differences in all these graphs up to that point.
00:24:40.800 --> 00:24:46.000
It is worth trying some of these out in your calculator, if you want to try graphing some of these in your calculator,
00:24:46.000 --> 00:24:52.500
just throw these values of α and β into the density function, and put the graphs on your calculator,
00:24:52.500 --> 00:24:54.300
and see what kinds of shapes you get.
00:24:54.300 --> 00:24:58.600
You get quite a lot of riots, it is kind of fun.
00:24:58.600 --> 00:25:04.200
In example 3 here, we are going to make a connection to a previous distribution that
00:25:04.200 --> 00:25:07.100
we learn which was the uniform distribution.
00:25:07.100 --> 00:25:11.200
It turns out that it is a special case of the β distribution.
00:25:11.200 --> 00:25:14.800
I did a whole video on the uniform distribution.
00:25:14.800 --> 00:25:20.900
If you do not remember the definition of the uniform distribution, or if you do not know what is it all,
00:25:20.900 --> 00:25:25.100
you can go back and look at the previous video covering the uniform distribution.
00:25:25.100 --> 00:25:28.500
You will see lots of information about the uniform distribution.
00:25:28.500 --> 00:25:31.200
You just need a quick refresher on the uniform distribution.
00:25:31.200 --> 00:25:39.300
I will remind you that the uniform distribution, the density function is just F of Y
00:25:39.300 --> 00:25:48.500
is always equal to 1/θ2 - θ1, where θ1 and θ2 are constants here.
00:25:48.500 --> 00:25:53.700
Those tripe lines there, the triple = means it is always equal to something.
00:25:53.700 --> 00:26:00.600
It is not varying at all, it is equal to this constant the whole way.
00:26:00.600 --> 00:26:09.400
That is where Y varies between θ1 and θ2.
00:26:09.400 --> 00:26:14.800
What we are going to do now is show how the β distribution, if you choose the right parameters
00:26:14.800 --> 00:26:18.000
turns into the uniform distribution.
00:26:18.000 --> 00:26:21.600
That was the uniform distribution that I showed you up above.
00:26:21.600 --> 00:26:28.800
Now, let me show you the formula for the β distribution which is much more complicated.
00:26:28.800 --> 00:26:54.700
F of Y is equal to Y ^Α -1 × 1 - Y ^β -1 ÷ B of Α β and that goes from Y between 0 and 1.
00:26:54.700 --> 00:27:01.200
I want to show you how you can choose the right parameters and turn the β distribution into the uniform distribution.
00:27:01.200 --> 00:27:09.700
I'm going to choose my Α is equal to 1 and my β is also equal to 1.
00:27:09.700 --> 00:27:11.500
Let us see how that worked out.
00:27:11.500 --> 00:27:16.700
First of all, let us find the constant value B of Α β.
00:27:16.700 --> 00:27:23.200
I'm going to use the relationship with the Gamma function, in order to figure that out.
00:27:23.200 --> 00:27:32.700
That is Γ of 1 × Γ of 1 ÷ Γ of 1 + 1.
00:27:32.700 --> 00:27:36.100
This is the relationship between the β function and the Gamma function,
00:27:36.100 --> 00:27:39.000
that I mentioned back in one of the earlier slides in the lecture.
00:27:39.000 --> 00:27:47.100
If you do not remember this, just go back and check your earlier slide and you will see the β and Γ relationship.
00:27:47.100 --> 00:27:55.900
Also, remember here that Γ of N is N -1!, when N is a whole number.
00:27:55.900 --> 00:28:04.200
This is 0! × 0! ÷ Γ of 2 would be 1!.
00:28:04.200 --> 00:28:10.300
Of course, all those factorials, 0 and 1 are just 1, this is all just 1.
00:28:10.300 --> 00:28:17.900
The β and density function is F of Y, that denominator is now going to be 1, we just work that out.
00:28:17.900 --> 00:28:29.700
Y ^α -1, α is 1 so α-1 is 0 × 1 - Y ^β -1 is also 0.
00:28:29.700 --> 00:28:33.900
And, that just comes out to be 1.
00:28:33.900 --> 00:28:37.100
It looks like it is going to be constant.
00:28:37.100 --> 00:28:42.900
Notice, by the way that, if it is constant then I can put three lines there,
00:28:42.900 --> 00:28:58.000
that is equal to 1/1 -0 that is 1/θ2 – θ1, where θ1 is 0 and θ2 is 1.
00:28:58.000 --> 00:29:17.000
What I found is that, I got the same density function for the β distribution as I would have gotten for the uniform distribution.
00:29:17.000 --> 00:29:29.000
This is the same as the uniform distribution.
00:29:29.000 --> 00:29:35.500
I discovered the uniform distribution as a special distinguished member of the β family.
00:29:35.500 --> 00:29:43.400
If you choose your α and β right, the β distribution just turns into the uniform distribution.
00:29:43.400 --> 00:29:51.000
Let me recap that, first of all, I recalled the uniform distribution and we did have a whole lecture on the uniform distribution.
00:29:51.000 --> 00:29:54.400
You can check the video on that, if you do not really remember how that worked out.
00:29:54.400 --> 00:29:59.800
The idea is you take Y between two values, θ1 and θ2.
00:29:59.800 --> 00:30:04.300
F of Y is just 1 ÷ θ2 – θ1.
00:30:04.300 --> 00:30:09.500
It is a constant distribution, there is no Y term appearing in there.
00:30:09.500 --> 00:30:14.000
And then, I reminded myself of the density function for the β distribution.
00:30:14.000 --> 00:30:17.100
That is our density function for the β distribution.
00:30:17.100 --> 00:30:22.700
I picked good values in the parameters, I pick α to be 1 and β to be 1.
00:30:22.700 --> 00:30:24.700
Then, I just plug those in.
00:30:24.700 --> 00:30:32.200
First, I had to calculate B of Α β and I turn that into an expression using Gamma functions.
00:30:32.200 --> 00:30:35.600
In turn, the Gamma functions turn into factorial.
00:30:35.600 --> 00:30:42.800
It just reduced down to 1 which why my denominator here was 1, that is where that one came from.
00:30:42.800 --> 00:30:48.400
And then, I plugged in my values of Α and β into the exponent, I got 0 is for the exponents,
00:30:48.400 --> 00:30:53.800
which why everything just disintegrated into big old 1 here.
00:30:53.800 --> 00:30:59.400
We just got that constant distribution F of Y is equal to 1, all the way across,
00:30:59.400 --> 00:31:05.700
which is the same as the uniform distribution on the interval 0-1.
00:31:05.700 --> 00:31:14.700
If you take θ1 = 0 and θ2 = 1, then we get the uniform distribution of 1 on that interval.
00:31:14.700 --> 00:31:22.300
The uniform distribution, it turns out, is a special case of the β distribution.
00:31:22.300 --> 00:31:28.700
In example 4, we are going to look at the triangular distribution F of Y = 2Y from 0 to 1,
00:31:28.700 --> 00:31:32.900
and show that that is also a special case of the β distribution.
00:31:32.900 --> 00:31:36.300
Let me just draw a quick graph of that triangular distribution.
00:31:36.300 --> 00:31:40.800
It is obvious why we are calling it the triangle distribution.
00:31:40.800 --> 00:31:49.100
Let me make my axis in black, I think that will show things a little better.
00:31:49.100 --> 00:31:52.500
We are going from Y = 0 to Y = 1 here.
00:31:52.500 --> 00:31:57.300
You always do that with the β distribution, it always goes from 0 to 1.
00:31:57.300 --> 00:32:03.600
F of Y = 2Y, that is just a straight line is not it.
00:32:03.600 --> 00:32:12.400
Probably, a little bit steeper than that, let me make that a little bit steeper.
00:32:12.400 --> 00:32:24.100
F of Y = 2Y, and then, the challenge here is to recognize that as a special case of the β distribution.
00:32:24.100 --> 00:32:28.400
Let me remind you of the density function for the β distribution.
00:32:28.400 --> 00:32:34.000
And then, we will take a look at it and see if we can make it match what we have here.
00:32:34.000 --> 00:32:47.500
For β distribution, F of Y is equal to Y ^Α -1 × 1 – Y ^β -1 ÷ B of Α, β.
00:32:47.500 --> 00:32:57.800
I want that to match 2Y, and I think what I want to do there is I want to pick Α equal to 2.
00:32:57.800 --> 00:33:01.000
That will make the exponent on the Y match.
00:33:01.000 --> 00:33:03.500
That 1 - Y does not really seem to match.
00:33:03.500 --> 00:33:09.600
In order to make that dropout, I'm going to take my β equal to 1.
00:33:09.600 --> 00:33:12.000
Let us plug those in and see how it works out.
00:33:12.000 --> 00:33:20.100
B of Α β, I could use the original integral definition to calculate that but I'm fond of the relationship
00:33:20.100 --> 00:33:23.100
between the β function and the Gamma function.
00:33:23.100 --> 00:33:24.400
I’m going to use that.
00:33:24.400 --> 00:33:33.100
It is Γ of Α × Γ of β ÷ Γ of α + β.
00:33:33.100 --> 00:33:37.200
Those are the relationship that we had back on one of the earlier slides in this lecture.
00:33:37.200 --> 00:33:40.100
You can look that up, if you do not remember it.
00:33:40.100 --> 00:33:47.800
Γ of Α is Γ of 2 × Γ of 1, β is 1 here.
00:33:47.800 --> 00:33:51.200
Γ of 2 + 1 is Γ of 3.
00:33:51.200 --> 00:33:56.100
Now, let us remember that the Gamma function is just a sort of
00:33:56.100 --> 00:34:01.300
a generalized version of the factorial function, except it shifted over by 1.
00:34:01.300 --> 00:34:07.600
This is 1! × 0! ÷ 2!.
00:34:07.600 --> 00:34:17.500
I’m shifting everything back by 1 that is because Γ of N is equal to N -1!, for whole numbers there.
00:34:17.500 --> 00:34:26.300
This is easy to solve, 1! Is 1, 0! Is 1, and 2! Is 2, that is ½.
00:34:26.300 --> 00:34:33.800
F of Y, now I know what my constant is, it is ½, is Y ^Α – 1.
00:34:33.800 --> 00:34:43.300
That is Y¹, 1 - Y ^β – 1, that is 1 - Y = 0 ÷ ½.
00:34:43.300 --> 00:34:48.700
That simplifies, the 1 - Y = 0 is just 1, it goes away.
00:34:48.700 --> 00:35:01.000
What we get here is Y by itself, but then, ÷ ½ is the same as multiplying by 2.
00:35:01.000 --> 00:35:06.900
It is 2Y, and of course Y is trapped between 0 and 1 here.
00:35:06.900 --> 00:35:19.100
That is very encouraging because that is the distribution, that is the density function that we are looking at.
00:35:19.100 --> 00:35:30.500
We started out with F of Y is 2Y, we found that to occur, if we pick the right parameters in the β distribution.
00:35:30.500 --> 00:35:35.900
Let me recap here, we started out wanting F of Y is equal 2Y.
00:35:35.900 --> 00:35:45.400
I wrote down the density function for the β distribution, Y ^α -1 1- Y ^β – 1/B of α, β.
00:35:45.400 --> 00:35:50.100
I’m trying to make it match to Y, I kind of looked at this exponent α -1.
00:35:50.100 --> 00:35:55.700
And I said, how can I make that be equal to 1, that will work if α is equal to 2.
00:35:55.700 --> 00:35:59.800
That 1 - Y really is not represented over here.
00:35:59.800 --> 00:36:07.500
The 1 - Y has a power of 0 here, I will take β equal 1 to make that work out.
00:36:07.500 --> 00:36:11.100
And then, I had to calculate the constant B of Α, β.
00:36:11.100 --> 00:36:14.900
I did that, by converting that into Gamma functions, according to the formula
00:36:14.900 --> 00:36:17.800
that we had on the earlier slides for this lecture.
00:36:17.800 --> 00:36:22.800
It is Γ of α × Γ of β ÷ Γ of α + β.
00:36:22.800 --> 00:36:30.400
If you drop the values of α and β data in there, then you will get something that we can simplify into factorials.
00:36:30.400 --> 00:36:37.300
Remember, there is a relationship between the Gamma function and the factorial function.
00:36:37.300 --> 00:36:44.300
Once, you simplified it into factorials, it simplifies quite nicely down into a fraction.
00:36:44.300 --> 00:36:51.100
The F of Y, if I fill in my α is 2 then I get Y¹.
00:36:51.100 --> 00:36:56.700
Β is 1 so I get 1 – Y = 0.
00:36:56.700 --> 00:37:01.200
That Y ÷ /2 which is 2Y.
00:37:01.200 --> 00:37:08.900
2Y from Y going from 0 to 1 is exactly the density function that we started with.
00:37:08.900 --> 00:37:21.200
It does match what we are given, we do achieve this triangular distribution as a special case of the β distribution.
00:37:21.200 --> 00:37:24.400
Let me show you quickly why it is called the triangular distribution.
00:37:24.400 --> 00:37:30.800
If you look at all the area there, the area is a triangle.
00:37:30.800 --> 00:37:41.200
The density is sort of spread out over a triangle there, which is why we call it the triangular distribution.
00:37:41.200 --> 00:37:49.300
In example 5, we got morning commute, maybe this is you driving to work in the morning or driving to school in the morning.
00:37:49.300 --> 00:37:57.600
It is a random variable, apparently, that has a β distribution with α and β both being 2.
00:37:57.600 --> 00:38:00.100
This is measuring your commune in hour.
00:38:00.100 --> 00:38:06.600
Remember, Y is always between 0 and 1 in the β distribution.
00:38:06.600 --> 00:38:13.300
I guess that means that your commute could be 0 or it could be up to an hour.
00:38:13.300 --> 00:38:18.300
In part A, we are going to find the chance that it will take longer than 30 minutes.
00:38:18.300 --> 00:38:24.100
What we will actually be looking for there is the probability that Y is bigger than or equal
00:38:24.100 --> 00:38:30.000
to 30 minutes is ½ an hour, it is bigger than or equal to ½.
00:38:30.000 --> 00:38:36.500
In part B, apparently, you have a rage level which is a function of how long your commute is.
00:38:36.500 --> 00:38:44.100
You want to find the expected value of rage that you will arrive at work tomorrow.
00:38:44.100 --> 00:38:45.700
Let us work this out.
00:38:45.700 --> 00:38:53.600
The first thing I'm going to do is try to identify the density function for this distribution.
00:38:53.600 --> 00:39:08.000
We got F of Y, the generic density function for the β distribution is Y ^α – 1 1- Y ^β – 1 ÷ B of Α, β.
00:39:08.000 --> 00:39:15.600
Let me go ahead and find the value of that constant, B of Α, β.
00:39:15.600 --> 00:39:23.700
I think the easiest way to calculate these, if you have whole numbers, is to convert it into Γ.
00:39:23.700 --> 00:39:34.500
Let me convert that into Γ of Α × Γ of β ÷ Γ of α + β.
00:39:34.500 --> 00:39:38.100
That is a formula that we learn back on one of the first slides in this lecture.
00:39:38.100 --> 00:39:45.100
In this case, we got Γ of 2 × Γ of 2 ÷ Γ of 4.
00:39:45.100 --> 00:39:49.400
Because α and β are both 2, that was given in the problem.
00:39:49.400 --> 00:39:55.000
This is 1! × 1!.
00:39:55.000 --> 00:39:57.700
Γ of 4 would be 3!.
00:39:57.700 --> 00:40:06.500
Remember, a part of the Gamma function is, it sort of the generalization of the factorial function but it is offset by 1.
00:40:06.500 --> 00:40:15.300
Γ of 4 is 3!, 1! Is 1, 3! Is 6, this is 1/6 here.
00:40:15.300 --> 00:40:25.100
This F of Y is Y ^α – 1, that is Y¹, 1 - Y ^β -1 is also 1.
00:40:25.100 --> 00:40:28.700
And now, we know that we are dividing it by 1/6.
00:40:28.700 --> 00:40:39.300
This simplifies down into 6Y × 1 - Y is Y - Y².
00:40:39.300 --> 00:40:45.200
I can go ahead and distribute that, 6 Y - 6 Y².
00:40:45.200 --> 00:40:49.700
That is my density function for this distribution.
00:40:49.700 --> 00:40:52.800
We are going to go ahead and use that to solve these two problems.
00:40:52.800 --> 00:41:00.700
I'm going to jump over onto next slide and use that density function to solve these two problems.
00:41:00.700 --> 00:41:03.400
But, let me recap where these came from.
00:41:03.400 --> 00:41:08.100
First, I was writing out the generic density function for the β distribution.
00:41:08.100 --> 00:41:09.900
There is the formula there.
00:41:09.900 --> 00:41:13.500
I have to figure out what B of Α, β was, I did that down below.
00:41:13.500 --> 00:41:21.600
I converted that into a bunch of Γ because I know easily how to solve the Gamma function,
00:41:21.600 --> 00:41:23.600
when you have a whole number in there.
00:41:23.600 --> 00:41:34.600
It is just the corresponding factorial, except you have to shift it down by 1 to plug the number into factorial.
00:41:34.600 --> 00:41:38.500
That is why B of Α, β turns out to be 1/6.
00:41:38.500 --> 00:41:49.700
I plugged that into my density function.
00:41:49.700 --> 00:41:53.300
I also plugged in the values of Α and β.
00:41:53.300 --> 00:42:01.600
2 -1 is 1, and I got 6 × Y - Y², that is the density function that I'm going to work with,
00:42:01.600 --> 00:42:05.700
in order to solve these two problems.
00:42:05.700 --> 00:42:09.500
On the next page, I’m not going to copy the problems because I need the space.
00:42:09.500 --> 00:42:13.400
We are going to solve the probability that Y is greater than ½.
00:42:13.400 --> 00:42:17.200
And then, we are going to find the expected value of R.
00:42:17.200 --> 00:42:25.500
Those are the two things that we are going to solve on the next page there.
00:42:25.500 --> 00:42:29.300
We are still working on example 5, we have the setup on the previous side.
00:42:29.300 --> 00:42:35.900
What we figured out was the density function is equal to 6 × Y - Y².
00:42:35.900 --> 00:42:42.200
For part A, we want to find the probability that your commute will be longer than 30 minutes.
00:42:42.200 --> 00:42:50.000
The probability that Y will be bigger than or equal to, we are measuring things in terms of hours.
00:42:50.000 --> 00:42:56.600
I converted 30 minutes into an hour, it is ½, that is the integral.
00:42:56.600 --> 00:43:09.200
Our range is from 0 to 1, ½ to 1 because you want it bigger than ½ of 6Y -6Y² DY.
00:43:09.200 --> 00:43:13.900
We have a little calculus problem to solve and it is not a hard one at all.
00:43:13.900 --> 00:43:21.800
Let us see, integral of 6Y is 3Y².
00:43:21.800 --> 00:43:27.000
The integral of 6Y² is 2Y³.
00:43:27.000 --> 00:43:34.600
We want to integrate that, to evaluate that from Y = ½ to Y = 1.
00:43:34.600 --> 00:43:45.800
That is 3 -2 -3 × ½² is ¾.
00:43:45.800 --> 00:43:51.400
+ 2 × ½³, ½³ is 1/8.
00:43:51.400 --> 00:43:54.800
2 × 1/8 is ¼.
00:43:54.800 --> 00:44:01.100
We get ¼ there.
00:44:01.100 --> 00:44:07.000
3 -2 is 1 - ¾ is ¼.
00:44:07.000 --> 00:44:20.700
This is ¼ + ¼, and that is ½, that is the probability that you are going to spend more than 30 minutes in traffic tomorrow morning.
00:44:20.700 --> 00:44:31.900
For part B, we had this rage level, your rage level was R is equal to Y² + 2Y + 1.
00:44:31.900 --> 00:44:35.300
What we want to calculate there was your expected rage level.
00:44:35.300 --> 00:44:38.900
How angry you expected to be, when you get to work?
00:44:38.900 --> 00:44:43.200
What we will use heavily here is the linearity of expectation.
00:44:43.200 --> 00:44:53.100
The expected value of R is the expected value Y² + 2Y + 1.
00:44:53.100 --> 00:45:04.100
But that is equal to the expected value of Y², this is linearity now, + 2 × the expected value of Y +,
00:45:04.100 --> 00:45:10.200
We can say that the expected value of 1 is just going to be 1.
00:45:10.200 --> 00:45:14.500
We need to find expected value of Y and Y².
00:45:14.500 --> 00:45:20.800
In order to figure out those, I'm going to remember what we learned on one of the very early slide.
00:45:20.800 --> 00:45:26.200
You can flip back and I think the slide was called key properties of β distribution.
00:45:26.200 --> 00:45:34.900
What we learn was that the expected value of Y was equal to Α ÷ Α + β.
00:45:34.900 --> 00:45:37.900
In this case, α and β are both 2.
00:45:37.900 --> 00:45:44.800
This is 2 ÷ 2 + 2, that is 2 ÷ 4 is ½.
00:45:44.800 --> 00:45:49.600
That was the easy one, the expected value of Y² is a little trickier.
00:45:49.600 --> 00:45:52.500
What we are going to do is use the variance.
00:45:52.500 --> 00:45:58.100
Σ², and that was kind of complicated formula, let me remind you what it was.
00:45:58.100 --> 00:46:01.300
This is coming from, I think it was the third slide in this lecture,.
00:46:01.300 --> 00:46:03.500
It was called key properties of the β distribution.
00:46:03.500 --> 00:46:10.700
(α + β)² × α + β + 1.
00:46:10.700 --> 00:46:13.400
We are going to work that out, α × β.
00:46:13.400 --> 00:46:19.500
Α and β are both 2, that was given to us on the previous slide.
00:46:19.500 --> 00:46:29.200
2 × 2 is 4 ÷ 2 + 2² is 16, and then 2 + 2 + 1 is 5.
00:46:29.200 --> 00:46:36.800
This 4 ÷ 16 is ¼ × 1/5 is 1/20.
00:46:36.800 --> 00:46:42.200
That is not the expected value of Y², that is the variance.
00:46:42.200 --> 00:46:46.100
Let me remind you how we use to calculate the variance.
00:46:46.100 --> 00:46:54.600
That is the expected value of Y² - the expected value of (Y)².
00:46:54.600 --> 00:47:00.700
What we can do is, we can use this to solve for the expected value of Y².
00:47:00.700 --> 00:47:08.300
This is a little old trick in probability is, if you know the variance, you can sort of reverse engineer for E of Y².
00:47:08.300 --> 00:47:20.200
E of Y² is equal to 1/20 + (E of Y)².
00:47:20.200 --> 00:47:25.400
We can write that as 1/20, we figure out what E of Y is, that is ½.
00:47:25.400 --> 00:47:44.000
We figure that out up above, ½², this is 1/20 + ¼ which is my common denominator, there is 20, 1/20 + ¼ is 5/20.
00:47:44.000 --> 00:47:50.600
I get here 6/20 and that reduces down to 3/10.
00:47:50.600 --> 00:47:57.800
My rage level, I decomposed it into E of Y² + 2E of Y + E of 1.
00:47:57.800 --> 00:48:10.600
Now, I can solve that, E of R is E of Y² + 2 E of Y + 1.
00:48:10.600 --> 00:48:16.000
I can solve that E of Y² is 3/10.
00:48:16.000 --> 00:48:26.800
2 E of Y is 2 × ½, I should have written E of 1 up above, here is E of 1.
00:48:26.800 --> 00:48:31.800
E of 1 is just 1 because that is the expected value of a constant.
00:48:31.800 --> 00:48:36.500
A constant is always going to be a constant, it is 1.
00:48:36.500 --> 00:48:47.000
This is 3/10 + 1 + 1, 3/10 + 2 is 23/10.
00:48:47.000 --> 00:48:50.600
All the Y is known as 2.3.
00:48:50.600 --> 00:48:54.600
I do not know what the units are there, I’m just going to leave them.
00:48:54.600 --> 00:48:58.900
I guess 2.3 range units is what you are going to carry into work,
00:48:58.900 --> 00:49:05.400
or at least the expected value of your rage as you could work tomorrow morning.
00:49:05.400 --> 00:49:09.700
2.3 is the expected value there.
00:49:09.700 --> 00:49:13.600
That answers both things that we were looking for here in this problem.
00:49:13.600 --> 00:49:16.700
Let me remind you where everything came from.
00:49:16.700 --> 00:49:21.000
This F of Y, this density function is something we figure out on the previous side.
00:49:21.000 --> 00:49:25.500
You can check back on the previous idea and see all the steps calculating that.
00:49:25.500 --> 00:49:31.300
In part A, we had to find the probability that it will take longer than 30 minutes to drive to work.
00:49:31.300 --> 00:49:36.900
30 minutes is ½ an hour and we want it to be longer than 30 minutes.
00:49:36.900 --> 00:49:44.800
We are going to integrate from ½ to 1, the density function, and that turns out to be a fairly straightforward integral.
00:49:44.800 --> 00:49:48.800
I’m going to keep the fractions straight.
00:49:48.800 --> 00:49:53.100
We get that the probability there is exactly ½.
00:49:53.100 --> 00:50:01.400
Half the time your commute will be longer than 30 minutes, half the time it would be shorter than 30 minutes.
00:50:01.400 --> 00:50:08.200
In part B, we had to find the expected value of your rage level, where it is defined like this.
00:50:08.200 --> 00:50:13.400
The expected value decomposes into these three terms.
00:50:13.400 --> 00:50:18.700
That is using linearity of expectation, very useful for this kind of problem.
00:50:18.700 --> 00:50:27.100
The expected value of Y, I am looking at the formulas for mean and variance on earlier slide from this talk.
00:50:27.100 --> 00:50:32.000
If you go back, just scroll up and you will see a slide called key properties of β distribution,
00:50:32.000 --> 00:50:36.300
and that is where I got these formulas, these complicated formulas using α and β.
00:50:36.300 --> 00:50:42.300
But then, I just plug in Α = 2 and β = 2, that was given to me in the problem.
00:50:42.300 --> 00:50:47.700
And, simplify those down into fractions ½ and 1/20.
00:50:47.700 --> 00:50:53.500
However, this was the variance, it was not the expected value of Y² directly.
00:50:53.500 --> 00:51:02.300
Remember, the old way to calculate variance is to find the expected value of Y² - the expected value of (Y)².
00:51:02.300 --> 00:51:08.800
What we can do is reverse engineer this, in order to solve for the term we want.
00:51:08.800 --> 00:51:12.800
Here is it, the term we want is the expected value of Y².
00:51:12.800 --> 00:51:20.700
I bring this E of Y² over to the other side and that is where I get 1/20 + E of Y².
00:51:20.700 --> 00:51:25.900
And then, this half right here is that half right there.
00:51:25.900 --> 00:51:28.200
That is where that half comes from.
00:51:28.200 --> 00:51:32.400
That is some more fractions, doing little simplification of the fractions.
00:51:32.400 --> 00:51:36.000
We get that E of Y² is exactly 3/10.
00:51:36.000 --> 00:51:49.500
I drop that into our expected value for R, and then this ½ is also that ½ from the E of Y.
00:51:49.500 --> 00:51:53.400
The expected value of 1 is always 1 because 1 is constant.
00:51:53.400 --> 00:51:56.100
You expect it to have its value.
00:51:56.100 --> 00:52:04.400
And then, simplifying down 3/10 + 1 + 1 is 23/10 or 2.3.
00:52:04.400 --> 00:52:07.800
That wraps up this lecture on the β distribution.
00:52:07.800 --> 00:52:11.100
This was the last of the continuous distributions.
00:52:11.100 --> 00:52:17.200
We worked through the uniform distribution, and the normal distribution, and the Gamma distribution.
00:52:17.200 --> 00:52:21.700
The Γ, of course, includes the exponential and the Chi square distribution.
00:52:21.700 --> 00:52:23.800
We had one big lecture on all of those.
00:52:23.800 --> 00:52:26.700
Finally, we got the β distribution.
00:52:26.700 --> 00:52:30.700
You are supposed to be an expert now on the continuous distributions.
00:52:30.700 --> 00:52:34.800
I got one more lecture in this chapter, it is going to cover moment generating functions.
00:52:34.800 --> 00:52:38.400
That is what you will see, if you stick around for the next lecture.
00:52:38.400 --> 00:52:43.100
In the meantime, you have been watching the probability lectures here on www.educator.com.
00:52:43.100 --> 00:52:45.000
My name is Will Murray, thank you for joining me today, bye.